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I see two ways to think about it...
Draw out a table of all possible sums with each possible die value across the columns and each down the rows.
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Now the first way to solve is to notice there are five(5) sums of 6.
Of these five sums, two(2) have one die that is 2:
Pr(one die is 2| sum is 6) = 2/5.
Another way is to apply Bayes' Theorem to the problem:
P(A|B) = P(B|A)*P(A) / P(B)
where
P(A) = P(one die is 2) = 11/36
P(B) = P(sum of dice is 6) = 5/36
P(B|A) = P(sum of dice is 6 given one die is a 2) = 2/11
P(A|B) = P(one die is 2 given sum is 6) = (2/11)(11/36) / (5/36) = 2/5
